# Parametric equation second derivative proof

Below is the question:

If $$x=\sin t$$ and $$y=\cos 2t$$, find $$\frac{dy}{dx}$$ in terms of $$x$$ and prove that $$\frac{d^2y}{dx^2} + 4 =0$$.

I found $$\frac{dy}{dx}$$ first (using the chain rule, since what they have given us is parametric equations) and got $$\frac{dy}{dx} = -\frac{2\sin 2t}{\cos t}$$ and then I derived that again to get the second derivative, but was not able to prove it.

For the second derivative I got $$\frac{d^2y}{dx^2} = -\frac{4cost}{\sin t} - \frac{4\cos 2t}{\cos t}.$$ After this I tried to bring the two together getting this: $$\frac{-4 \cos t^{2} - 4\sin t\ cos 2t}{\sin t\cos t}$$

But then did not know how to get it to $$-4$$ (considering $$-4+4=0$$).

Can someone help me on how to solve this? As well as point out what I have done wrong?

Thank you.

find $$\frac{dy}{dx}$$ in terms of $$x$$

Hint: use the expansion of $$\sin 2t = 2\sin t\cos t$$

You have: $$\frac{dy}{dx} = -\frac{2\sin{2t}}{\cos t} = -\frac{4\sin t\cos t}{\cos t} = -4\sin t$$ $$\Rightarrow \frac{dy}{dx} = -4x$$

The rest should be straightforward.

Note: I think the reason you did not get to the same result is because you have not used the quotient rule for the second-order derivative.

• So simplify dy/dx before deriving it again! Makes so much sense thank you – user639649 Apr 10 at 1:05
• @user639649 also see the note. You probably didn't use the quotient rule, else you should have got the same answer by your method as well. – user1952500 Apr 10 at 1:07
• But can I not use product rule regardless? By making the bottom values to the power of -n ? – user639649 Apr 10 at 1:07
• @user639649 the product rule is fine, and it seems intuitive because the relation is similar to normal multiplication. However you should use it carefully and realize that you are working with derivatives. – user1952500 Apr 10 at 1:09
• Okay, Ill try again using my initial method just with quotient rule, thank you! – user639649 Apr 10 at 1:09